Question

The length of the perpendicular from the origin to a line is $7$ and the line makes angle of ${150}^o$. Which positive direction of y-axis. Find the equation of line.

Answer 1

Given,

angle made with the positive y-axis$={150}^{\circ }$

$\therefore$ Angle made with positive x-axis$={60}^{\circ }$ and ${180}^{\circ }-{60}^{\circ }={120}^{\circ }$

Let intercepts made by the line on x & y axis be a & b respectively.

$\therefore$ Co-ordinates are $A(0,0)$ & $B(0,b)$.

$\therefore$Slope$=\tan {120}^{\circ }$

$\Rightarrow \frac{b-0}{0-a}=-\sqrt{3}$

$\Rightarrow b=+\sqrt{3}a$--------------(1)

or,

Slope$=\tan {60}^{\circ }$

$\Rightarrow \frac{b-0}{0-a}=\sqrt{3}$

$\Rightarrow b=-\sqrt{3}a$---------------(2)

$\therefore$ Equation of the lines will be

$y-0=\frac{b}{-a}(x-a)$ ; $y=\sqrt{3}(x-a)$----------------(4)

$\Rightarrow y=-\sqrt{3}(x-a)$----------(3)

Also given, that the distance from origin is 7 units.

$d_{origin}=7$

$\Rightarrow \mid \frac{Ax+By+C}{\sqrt{A^2+B^2}}\mid =7$

$\Rightarrow \frac{\mid \sqrt{3}\left(0\right)\pm 1\left(0\right)-a\sqrt{3}\mid }{\sqrt{3+1}}=7$

$\Rightarrow a\sqrt{3}=\pm 14$

$\therefore$ Equation of lines are:

$\sqrt{3}x+y\pm 14=0$ and $\sqrt{3}x-y\pm 14=0$